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I was wondering if someone could please suggest a technique for solving the following system of ODEs:

$$x_1'=(1+2\cos2t)x_1 + (1-2\sin2t)x_2$$

$$x_2'=-(1+2\sin2t)x_1 + (1-2\cos2t)x_2$$

What I initially tried to do was differentiate the first equation to obtain an equation for $x_1''$ and then substitute expressions for $x_2$ and $x_2'$. This resulted in a second-order DE involving $x_1''$, $x_1'$, and $x_1$. But this equation was extremely complex in terms of its variable coefficients. I am thinking there must be a simpler approach. Thanks.

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1 Answer 1

I am not 100% sure if this works but I think you can reduce it to a linear equation:

$$ (1 + 2 \sin 2t) x_1' + (1 + 2\cos 2t) x_2' = (1 - 4 \sin^2 2t) x_2 + (1 - 4 \cos^2 2t) x_2 = -2 x_2$$

$$ (1 - 2\cos 2t) x_1' - (1 - 2 \sin 2t) x_2' = (1 + 4 \cos^2 2t) x_1 + (1 + 4 \sin^2 2t) x_1 = 6 x_1$$

Substituting back into the original equation:

$$ x_1' = -\frac{1}{2}(1 + 2\cos 2t)((1 + 2 \sin 2t) x_1' + (1 + 2\cos 2t) x_2') + \frac{1}{6} (1 - 2 \sin 2t)((1 - 2\cos 2t) x_1' - (1 - 2 \sin 2t) x_2') = \frac{1}{6} (\sin {2 t} (-8 x_1'+x_2'+2)-2 \cos {2 t} (4 x_1'+3 x_2'+3)-4 x_1' \sin {4 t}-5 x_2' \cos {4 t}-2 x_1'-7 x_2'-4) $$

$$x_2' = \frac{1}{6} (\sin{2 t} (12 x_1'+x_2'+6)+\cos{2 t} (-4 x_1'+6 x_2'+2)-4 x_1' \cos (4 t)+5 x_2' \sin{4 t}+12 x_1'+2)$$

I am assuming this can be solved. I'll leave this one posted while I think of some better ways.

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